Are chain complexes chain equivalent to free ones?

Question

Given a chain complex $A_bulletinmathrm{Ch(mathbf{Ab})}$, are there exist some chain complex $A'_bulletinmathrm{Ch(mathbf{Ab})}$ which is chain equivalent to $A_bullet$ such that $A'_p$ are all free abelian groups?

Yuta · Answer

I'm answering my own question.
No. A counter example is $A_bullet := A_0to A_1to A_2 := 0tomathbf{Z}/2mathbf{Z}to 0$.
Assume there exists a chain complex $A'_bullet$ of free abelian group with $varphi:A_bulletto A'_bullet, psi:A'_bulletto A_bullet$ and a homotopy $alpha$ between $psicircvarphi$ and $mathrm{id}_{A_bullet}$.
It lead to a contradiction Because

$phi_1:mathbf{Z}/2mathbf{Z}to B$ is a null map because $B$ is free and
the only $alpha_0$ and $alpha_1$ are null maps.

Are chain complexes chain equivalent to free ones?

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